I need to extract the (, ), - and numbers from 0 to 9 from the input field phone number.
For eg., for input data : (891)456-567A$rT67
Output should be (891)456-56767
can be done in below 2 steps -
v_ph_no = REG_REPLACE( in_ph_no,'[^0-9]','')` -- variable port that will replace everything but 0 to 9.
out_ph_no = '(' || substr(v_ph_no,1,3)||')' || substr (v_ph_no,4,3)||'-' || substr (v_ph_no,6) -- o/p port that will create ph no in desired format.
Explanation - first remove all unwanted characters and then format it in desired phone no format.
Related
I have Id and email in flat file, I am using expression transformation, created an output port as string data type, trying to pass the value as below
'[
{
"Id":"'||Id||'"
"email":"'||email||'"
}
]'
output:
"[
{
"Id":"2904"
"email":"k9#gmail.com"
}
]"
I don't need the " " at the start and end for square braces and no extra space should be there after line ends .
expected output:
[
{
"Id":"2904"
"email":"k9#gmail.com"
}
]
I will pass this string to the API webservice as http request. Kindly help with required format.
In exp transformation create below ports and logic.
--assuming input column is datacol
var_str1 = LTRIM(substr(datacol, substr(datacol,'output: '))) -- This will result everything after the word output:.
var_str2 = substr( var_str1 ,2) -- removes first double quote
var_str3 = substr( var_str2 ,-1) -- removes last double quote
out_str= var_str3
I did them in step by step process. if you are comfortable, pls combine all into one line.
EDIT
var_* - this denotes these are variable ports.
out_ - this output these are variable ports.
I'm building a string in oracle, where I get a number from a column and make it a 12 digit number with the LPad function, so the length of it is 12 now.
Example: LPad(nProjectNr,12,'0') and I get 000123856812 (for example).
Now I want to split this string in parts of 3 digit with a "\" as prefix, so that the result will look like this \000\123\856\812.
How can I archive this in a select statement, what function can accomplish this?
Assuming strings of 12 digits, regexp_replace could be a way:
select regexp_replace('000123856812', '(.{3})', '\\\1') from dual
The regexp matches sequences of 3 characters and adds a \ as a prefix
It is much easier to do this using TO_CHAR(number) with the proper format model. Suppose we use \ as the thousands separator.... (alas we can't start a format model with a thousands separator - not allowed in TO_CHAR - so we still need to concatenate a \ to the left):
See also edit below
select 123856812 as n,
'\' || to_char(123856812, 'FM000G000G000G000', 'nls_numeric_characters=.\') as str
from dual
;
N STR
--------- ----------------
123856812 \000\123\856\812
Without the FM format model modifier, TO_CHAR will add a leading space (placeholder for the sign, plus or minus). FM means "shortest possible string representation consistent with the model provided" - that is, in this case, no leading space.
Edit - it just crossed my mind that we can exploit TO_CHAR() even further and not need to concatenate the first \. The thousands separator, G, may not be the first character of the string, but the currency symbol, placeholder L, can!
select 123856812 as n,
to_char(123856812, 'FML000G000G000G000',
'nls_numeric_characters=.\, nls_currency=\') as str
from dual
;
SUBSTR returns a substring of a string passed as the first argument. You can specify where the substring starts and how many characters it should be.
Try
SELECT '\'||SUBSTR('000123856812', 1,3)||'\'||SUBSTR('000123856812', 4,3)||'\'||SUBSTR('000123856812', 7,3)||'\'||SUBSTR('000123856812', 10,3) FROM dual;
I am working on PL/SQL . The oracle password by developer is set such way
=> input word => converted to ascii => added 2 to each letter => converted back to word
ex: input password is "admin".
admin is splitted into characters/letters (a, d, m, i, n)
converted to ascii and added 2 and again converted to word
a=97 97+2 = 99 = c
d=100 100+2=102 = f
m=109 109+2=111 = o
i=105 105+2=107 = k
n=110 110+2=112 = p
what i did is
$pass=str_split('admin');
foreach($pass as $password){
$new_password[]=chr(ord($password)+2);
}
$final= $new_password[0].$new_password[1].$new_password[2].$new_password[3].$new_password[4]; //the values 0-4 is set manually
echo $final;
result: cfokp
But i could not get proper ans to run the result string on command and match the oracle password with the retrieved one.
Another way in SQL is to split the characters, add 2 to the ascii value, and aggregate the string.
Of course, it won't be faster than the TRANSLATE approach. But, for a single or small set of values it shouldn't matter much.
For example,
SQL> WITH data AS
2 (SELECT 'admin' str FROM dual
3 )
4 SELECT str, LISTAGG(CHR(ASCII(REGEXP_SUBSTR(str, '\w', 1, LEVEL)) + 2), '') WITHIN GROUP(
5 ORDER BY LEVEL) str_new
6 FROM data
7 CONNECT BY LEVEL <= LENGTH(str)
8 /
STR STR_NEW
------ -------
admin cfokp
SQL>
The above SQL does following important tasks:
Split the string into characters using REGEXP_SUBSTR and ROW GENERATOR technique
Add value 2 to the ascii value of each character.
Convert back the modified ascii into characters.
Aggregate the string using LISTAGG
This is probably easier to do with translate:
select translate('admin',
'abcdefghijklmnopqrstuvwxyz',
'cdefghijklmnopqrstuvwxyzab'
)
from dual;
I'm not sure what you want to do with "y" and "z". This maps them back to "a" and "b". You can extend this to upper case letters and other characters if you need.
In Oracle I want to check whether the string has "=' sign at the end. could you please let me know how to check it. If it has '=' sign at the end of string, I need to trailing that '=' sign.
for eg,
varStr VARCHAR2(20);
varStr = 'abcdef='; --needs to trailing '=' sign
I don't think you need "pattern matching" here. Just check if the last character is the =
where substr(varstr, -1, 1) = '='
substr when called with a negative position will work from the end of the string, so substr(varstr,-1,1) extracts the last character of the given string.
Use the REGEX_EXP function. I'm putting a sql command since you didn't specify on your question.:
select *
from someTable
where regexp_like( someField, '=$' );
The pattern $ means that the precedent character should be at the end of the string.
see it here on sql fiddle: http://sqlfiddle.com/#!4/d8afd/3
It seems that substr is the way to go, at lease with my sample data of about 400K address lines this returns 1043 entries that end in 'r' in an average of 0.2 seconds.
select count(*) from addrline where substr(text, -1, 1) = 'r';
On the other hand, the following returns the same results but takes 1.1 seconds.
select count(*) from addrline where regexp_like(text, 'r$' );
I'd like to know if there is a way to check if there is a comma , in the !field.Value.
I want to make these conversations:
10,5 -> 10,50
900 -> 900,00
To do that, I need to know if there is a comma in the field value and also how many characters are after the comma. Is it possible ?
Look at InStr(), Len(), and IIF(), I think they will get you what you want.
I don't have a way to test this where I am, but basically I think this expression will get you there:
=IIF(InStr(Fields!MyField.Value, ",") > 0,
Fields!MyField.Value & LEFT("000000", (-1 *(2 - (Len(Fields!MyField.Value) - InStr(Fields!MyField.Value, ","))))),
Fields!MyField.Value & ",00")
Here's the basic idea of the script:
If there is a comma in the field,
then add x number of 0s onto the end of the field
where x is 2 - (the length of the field - the position of the ',' in the string) * -1
else just return the field + ",00"